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34kurt
3 years ago
10

Why is it advantageous to have two Eyes?​

Physics
1 answer:
Andru [333]3 years ago
6 0

Answer:para enxergar

Explanation:

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Explain the expression “light travels at a speed”.
Anestetic [448]
Some things travel faster than the speed of light
7 0
3 years ago
What is the relation between celsius and kelvin​
Scilla [17]

Answer:

The Celcius and kelvin scale are related unit for unit. One degree unit on the Celcius scale is equivalent to one degree unit on the kelvin scale. The only difference between these two scales is the zero point.

7 0
3 years ago
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An 80 kg astronaut has gone outside his space capsule to do some repair work. Unfortunately, he forgot to lock his safety tether
Anna [14]

Answer:

the time taken by the astronaut to reach safety = 9.8 hr

Explanation:

The equation for intensity can be written as :

I = \frac{P}{A}

where :

\frac{I}{c}= \frac{F}{A}

Replacing that into the above previous equation; we have:

\frac{P}{Ac}=\frac{F}{A}

F = \frac{P}{c}

However ; the force needed to push the astronaut is as follows:

F = ma

where ;

m = mass of the  astronaut and a = its acceleration

we as well say;

\frac{P}{c} = ma

a = \frac{P}{mc}

Replacing P with 1000 W ; m with 80 kg and 3*10^{8} \  m/s for  c

Then; a = \frac{1000 \ W}{(80)(3.0*10^8)}

a = 4.2*10^{-8} \ m/s

It is also known that the battery will run for one hour and after which the battery on the laser will run out

Then to determine the change in the position after the first hour ; we have:

\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 \ h)^2

\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 *3600 s)^2

= 0.27 m

Furthermore, the final velocity of the astronaut is determined as:

v_1 = at_1

where ;

v_1 = final \ velocity

replacing t_1 = 1.0 \ h and a =  4.2*10^{-8} \ m/s; Then:

v_1 = (4.2*10^8 \ m/s * 1.0 \ h * \frac{ 3600\ s}{1.0 \ h})

v_1 =  1.51 *10^{-4} \ m/s

Also; when he drifted 5.0 m away from the capsule; the distance is far short of the 5 m but he still have 9 hours left of oxygen . In addition to that, he acceleration is also zero and the final velocity remains the same, so:

To find the final distance traveled by the astronaut ;we have:

\Delta x_2 = d - \Delta x_1

where;

\Delta x_2 = the final distance

d = total distance

So;

\Delta x_2 = 5 m - 0.27 m \\ \\ \Delta x_2 = 4.73 \ m

The time taken to reach the final distance can be calculated as:

t_2 = \frac{\Delta x_2 }{v_1}

where;

t_2 = is the  time to  reach the final distance

Replacing 4.73 for {\Delta x_2 } and  1.51*10^{-4} m/s for v_1

t_2 = \frac{4.73 \ m }{1.51*10^{-4} \ m/s}

t_2 = 31500 \ s (\frac{1.0 \ h}{3600 \ s} )

t_2 = 8.8 \ h

We knew the laser was operated for 1 hour; thus the total time taken by the astronaut to  reach the final distance is the sum of the time taken to reach the final distance and the operated time of the laser.

Hence ; the time taken by the astronaut to reach safety = 9.8 hr

8 0
3 years ago
A 19 kg solid disk of radius0.44 m is rotated about an
Vlad1618 [11]

Answer:

0.915 Nm

Explanation:

1 revolution = 2π rad

We can use the following equation of motion to find out the acceleration acting on the disk

\omega^2 - \omega_0^2 = 2\alpha\Delta \theta

where \omega v = 2.5 rad/s is the final angular velocity of the disk, [tex]\omega_0 = 0 rad/s is the initial velocity of the can when it starts from rest, \Delta \theta is the angular distance traveled, \alpha is the angular acceleration of the disk, which we care looking for:

2.5^2 - 0 = 2*\alpha*2\pi

\alpha = \frac{2.5^2}{2*2\pi} \approx 0.5 rad/s^2

The moment of inertia of the solid disk is:

I = \frac{1}{2}mR^2 = \frac{1}{2}19*0.44^2 = 1.8392 kgm^2

where m is the mass and R is the radius of the disk

The net torque applied is

T = \alpha*I = 0.5 * 1.8392 = 0.915 Nm

8 0
3 years ago
A body of mass 0.1kg tied by a string is rotating a vertical circle with a speed of 10 m/s of radius a 1 m . what is the tension
allochka39001 [22]

Answer:

The tension experienced in the string at the highest point is approximately 9.019 Newtons

Explanation:

The mass of the body tied to the string, m = 0.1 kg

The path of motion of the mass = Vertical circular motion

The speed of rotation of the stone, v = 10 m/s

The radius of the circular motion path, r = 1 m

The required information = The tension at the highest point, T

At the highest point, the tension, 'T', comprises of the centrifugal force, 'F', acting upwards in the string and the weight, 'W' of the body acting downwards

∴ T = F - W

The \ centrifugal \  force, \ F = \dfrac{m \times v^2}{r}

Weight, W = m × g

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

The centrifugal force, F = 0.1 kg × (10 m/s)²/(1 m) = 10 N

The weight of the mass tied to the string, W ≈ 0.1 kg × 9.81 m/s² = 0.981 N

From which we have;

T = 10 N - 0.981 N = 9.019 N

The tension experienced in the string at the highest point ≈ 9.019 Newtons

8 0
3 years ago
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